Question

The vertical distance between the masses is 20/3 m
. The time at which the distance becomes zero is
a) 2
b) √25
c) 45
d) √6s
​

Answer 1

Answer:

a- 2 second

The answer is explained above.

Answer 2

Answer:

Option (a) 2 is the correct answer.

Explanation:

Given in the question that, the vertical distance between the masses is $\frac{20}{3}$m.

Now ,as wee know that, acceleration = $\frac{driving \ force}{Total \ mass}$

Acceleration = $\frac{(m_1 - m_2)g}{m_1 + m_2}$ = $\frac{(2 - 1)10}{2 + 1}$ = $\frac{10}{3}$

[ Where from the question $m_1 = 2kg \ and\ m_2 = 1kg$]

And Displacement Equations for these Calculations:

       $S$= .$ut + \frac{1}{2} at^2$....(i)

Where:

s = displacement

u = initial velocity

a = acceleration

t = time

On putting the values  in (i) we get,

⇒ s = $ut + \frac{1}{2} at^2$

⇒$\frac{20}{3}$ = 0 + $\frac{1}{2} . \frac{10}{3} .t^2$

⇒20 = 5$t^2$

⇒ t = $\sqrt{\frac{20}{5} }$ = $\sqrt{4}$ = 2

Hence, the time at which the distance becomes zero is 2 sec.

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